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Home Backend Development PHP Tutorial php下载问题,下载后文件内容是空的,图片也没有预览

php下载问题,下载后文件内容是空的,图片也没有预览

Jun 23, 2016 pm 02:12 PM

有两个文件,a.php 和 b.php 
点击a.php中的链接,开始对文件下载:代码如下

这是a.php

<?php $file_dir = "D:/files";$file_name = $row[1]; //这是从数据库中查找出来的echo "<a href='./b.php?file_dir=" . $file_dir . "&file_name=" . $file_name . "'>" . $row[2] . "</a>"; //点击这个链接,下载的文件名是对的,就是文件里没有内容,如果是图片,图片没有预览?>
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这是b.php的全部代码
<?php	$file_name = $_GET['file_name'];   //文件名	$file_dir  = $_GET['file_dir'];      //文件路径		if(!file_exists($file_dir.$file_name)){		 echo "找不到 [" . $file_dir.$file_name . "] 文件";		 exit;	}else{		 $file=fopen($file_dir.$file_name,"r");			Header("Content-type: application/octet-stream");		Header("Accept-Ranges:bytes");		Header("Accept-Length:".filesize($file_dir.$file_name));		Header("Content-Disposition: attachment; filename=".$file_name);	}?>
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回复讨论(解决方案)

a.php 中是 $file_dir = "D:/files";
那么在 b.php 中的 $file_dir.$file_name 就缺少了路径符

你在 b.php 中只有 $file=fopen($file_dir.$file_name,"r"); 打开了文件,但并没有输出文件的内容

如果用 $file=fopen($file_dir.$file_name,"r"); 打开的文件是图片的话,将不能正确读出图片数据
应使用二进制方式打开:$file=fopen($file_dir.$file_name,"r b");

a.php 中是 $file_dir = "D:/files";
那么在 b.php 中的 $file_dir.$file_name 就缺少了路径符

你在 b.php 中只有 $file=fopen($file_dir.$file_name,"r"); 打开了文件,但并没有输出文件的内容

如果用 $file=fopen($file_dir.$file_name,"r"); 打开的文件是图片的话,将不能正确读出图片数据
应使用二进制方式打开:$file=fopen($file_dir.$file_name,"r b");

听懂了 但是我改成了这样 还是一样的结果 代码如下

<?php	$file_name = $_GET['file_name'];   //文件名	$file_dir  = $_GET['file_dir'];      //文件路径为当前目录		if(!file_exists($file_dir.$file_name)){		 echo "找不到 [" . $file_dir.$file_name . "] 文件";		 exit;	}else{        $file=fopen($file_dir.$file_name,"rb");	readfile($file);             Header("Content-type: application/octet-stream");        Header("Accept-Ranges:bytes");        Header("Accept-Length:".filesize($file_dir.$file_name));        Header("Content-Disposition: attachment; filename=".$file_name);	ob_clean(); 	flush();	exit; ?>
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哎哟 晕 代码里不能改字体颜色哦 重新

<?php    $file_name = $_GET['file_name'];   //文件名    $file_dir  = $_GET['file_dir'];      //文件路径为当前目录         if(!file_exists($file_dir.$file_name)){         echo "找不到 [" . $file_dir.$file_name . "] 文件";         exit;    }else{        $file=fopen($file_dir.$file_name,"rb");        readfile($file);              Header("Content-type: application/octet-stream");        Header("Accept-Ranges:bytes");        Header("Accept-Length:".filesize($file_dir.$file_name));        Header("Content-Disposition: attachment; filename=".$file_name);        ob_clean();         flush();        exit;?>
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本帖最后由 xuzuning 于 2013-06-08 16:08:01 编辑

    }else{        $file = $file_dir.$file_name;        Header("Content-type: application/octet-stream");        Header("Accept-Ranges:bytes");        Header("Accept-Length:".filesize($file));        Header("Content-Disposition: attachment; filename=".$file_name);        readfile($file);    }
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成功了,首先要谢谢版主,版主说的非常好,然后必须分享一下:
是下载压缩文件的,包括压缩文件其中的文件,代码也不难

//打开文件		$file = fopen($file_dir.$file_name,"r");		//返回的文件类型		Header("Content-type: application/octet-stream");		//按照字节大小返回		Header("Accept-Ranges: bytes");		//返回文件的大小		Header("Accept-Length: ".filesize($file_dir.$file_name));		//这里对客户端的弹出对话框,对应的文件名		Header("Content-Disposition: attachment; filename=".$file_dir.$file_name);		//修改之前,一次性将数据传输给客户端		echo fread($file, filesize($file_dir.$file_name));		//修改之后,一次只传输1024个字节的数据给客户端		//向客户端回送数据		$buffer=1024;//		//判断文件是否读完		while (!feof($file)) {			//将文件读入内存			$file_data=fread($file,$buffer);			//每次向客户端回送1024个字节的数据			echo $file_data;		}		fclose($file);
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